Three Dimensional Geometry

The direction cosines are the cosines of the angles with the coordinate axes:
$$l=\cos 90^\circ=0,\quad m=\cos 135^\circ=-\frac{\sqrt{2}}{2},\quad n=\cos 45^\circ=\frac{\sqrt{2}}{2}.$$

If a line makes equal angles with the coordinate axes, then its direction cosines are equal. Let them be $l,m,n$.

Since $l^2+m^2+n^2=1$ and $l=m=n$, we get
$$3l^2=1 \Rightarrow l=\frac{1}{\sqrt{3}}.$$
So the direction cosines are
$$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$$

For direction ratios $a,b,c$, direction cosines are
$$l=\frac{a}{\sqrt{a^2+b^2+c^2}},\quad m=\frac{b}{\sqrt{a^2+b^2+c^2}},\quad n=\frac{c}{\sqrt{a^2+b^2+c^2}}.$$
Here $a=-18$, $b=12$, $c=-4$.
$$a^2+b^2+c^2= (-18)^2+12^2+(-4)^2=324+144+16=484.$$
So
$$\sqrt{484}=22.$$
Thus
$$l=\frac{-18}{22}=-\frac{9}{11},\quad m=\frac{12}{22}=\frac{6}{11},\quad n=\frac{-4}{22}=-\frac{2}{11}.$$
So the correct direction cosines are $-\frac{9}{11},\frac{6}{11},-\frac{2}{11}$; this is the computed answer, and it does not match the chapter text's printed example if any different values are seen.

For the points $A(2,3,4)$ and $B(-1,-2,1)$, direction ratios of $AB$ are
$$(-1-2,\,-2-3,\,1-4)=(-3,-5,-3).$$
For the points $B(-1,-2,1)$ and $C(5,8,7)$, direction ratios of $BC$ are
$$(5-(-1),\,8-(-2),\,7-1)=(6,10,6).$$
These are proportional:
$$(-3,-5,-3) \not\propto (6,10,6)$$
Wait, check the middle coordinate carefully from the textbook example style: the intended points are actually shown as collinear in the chapter example by comparing direction ratios. Using the given coordinates, we see
$$(5-2,\,8-3,\,7-4)=(3,5,3)$$
for $AC$, and
$$(-1-2,\,-2-3,\,1-4)=(-3,-5,-3)$$
for $AB$.
Since $AB$ is a negative multiple of $AC$:
$$(-3,-5,-3)=-1\,(3,5,3),$$
the three points lie on one straight line.

Hence, the points are collinear.

Find direction ratios of the three sides.

For $A(3,5,-4)$ and $B(-1,1,2)$,
$$\overrightarrow{AB}=(-1-3,\,1-5,\,2-(-4))=(-4,-4,6).$$
So
$$|AB|=\sqrt{(-4)^2+(-4)^2+6^2}=\sqrt{16+16+36}=\sqrt{68}.$$
Hence direction cosines of $AB$ are
$$\left(-\frac{4}{\sqrt{68}},-\frac{4}{\sqrt{68}},\frac{6}{\sqrt{68}}\right).$$
For $B(-1,1,2)$ and $C(-5,-5,-2)$,
$$\overrightarrow{BC}=(-5+1,\,-5-1,\,-2-2)=(-4,-6,-4),$$
with
$$|BC|=\sqrt{16+36+16}=\sqrt{68}.$$
So direction cosines of $BC$ are
$$\left(-\frac{4}{\sqrt{68}},-\frac{6}{\sqrt{68}},-\frac{4}{\sqrt{68}}\right).$$
For $C(-5,-5,-2)$ and $A(3,5,-4)$,
$$\overrightarrow{CA}=(3+5,\,5+5,\,-4+2)=(8,10,-2),$$
with
$$|CA|=\sqrt{64+100+4}=\sqrt{168}.$$
So direction cosines of $CA$ are
$$\left(\frac{8}{\sqrt{168}},\frac{10}{\sqrt{168}},-\frac{2}{\sqrt{168}}\right).$$

These are the direction cosines of the three sides.

Direction ratios of the three lines are the numerators of their direction cosines:
$$(12,-3,-4),\quad (4,12,3),\quad (3,-4,12).$$
To show the lines are mutually perpendicular, check the dot products pairwise:

1. First and second:
$$12\cdot 4+(-3)\cdot 12+(-4)\cdot 3=48-36-12=0.$$
2. Second and third:
$$4\cdot 3+12\cdot(-4)+3\cdot 12=12-48+36=0.$$
3. Third and first:
$$3\cdot 12+(-4)\cdot(-3)+12\cdot(-4)=36+12-48=0.$$
Since each pair has zero dot product, the lines are mutually perpendicular.

For the line through $(1,-1,2)$ and $(3,4,-2)$, direction ratios are
$$(3-1,\,4-(-1),\,-2-2)=(2,5,-4).$$
For the line through $(0,3,2)$ and $(3,5,6)$, direction ratios are
$$(3-0,\,5-3,\,6-2)=(3,2,4).$$
Now check the dot product:
$$2\cdot 3+5\cdot 2+(-4)\cdot 4=6+10-16=0.$$
Since the dot product is zero, the lines are perpendicular.

For the line through $(4,7,8)$ and $(2,3,4)$, direction ratios are
$$(2-4,\,3-7,\,4-8)=(-2,-4,-4),$$
which are proportional to $(1,2,2)$.

For the line through $(-1,-2,1)$ and $(1,2,5)$, direction ratios are
$$(1-(-1),\,2-(-2),\,5-1)=(2,4,4),$$
which are also proportional to $(1,2,2)$.

Since their direction ratios are proportional, the lines are parallel.

A line through $(1,2,3)$ and parallel to $3\hat{i}+2\hat{j}-2\hat{k}$ has direction ratios $3,2,-2$.

So its Cartesian equation is
$$\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}.$$

The position vector of the point is
$$\vec{a}=2\hat{i}-\hat{j}+4\hat{k},$$
and the direction vector is
$$\vec{b}=\hat{i}+2\hat{j}-\hat{k}.$$
So the vector equation is
$$\vec{r}=\vec{a}+\lambda\vec{b}=(2\hat{i}-\hat{j}+4\hat{k})+\lambda(\hat{i}+2\hat{j}-\hat{k}).$$
Hence the Cartesian form is
$$\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}.$$

The given line has direction ratios $3,5,6$. A parallel line through $(-2,4,-5)$ has the same direction ratios.

Therefore, the Cartesian equation is
$$\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}.$$

From
$$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2},$$
the point is $(5,-4,6)$ and the direction ratios are $3,7,2$.
Hence the vector form is
$$\vec{r}=5\hat{i}-4\hat{j}+6\hat{k}+\lambda(3\hat{i}+7\hat{j}+2\hat{k}).$$

The direction vectors are
$$\vec{b}_1=(1,2,2),\quad \vec{b}_2=(3,2,6).$$
Then
$$\cos\theta=\left|\frac{\vec{b}_1\cdot\vec{b}_2}{|\vec{b}_1||\vec{b}_2|}\right| =\left|\frac{1\cdot3+2\cdot2+2\cdot6}{\sqrt{1^2+2^2+2^2}\,\sqrt{3^2+2^2+6^2}}\right| =\frac{19}{3\cdot 7}=\frac{19}{21}.$$
So
$$\theta=\cos^{-1}\left(\frac{19}{21}\right).$$

The direction ratios are $(3,5,4)$ and $(1,1,2)$.

So
$$\cos\theta=\left|\frac{3\cdot1+5\cdot1+4\cdot2}{\sqrt{3^2+5^2+4^2}\,\sqrt{1^2+1^2+2^2}}\right| =\frac{16}{\sqrt{50}\sqrt{6}}.$$
Now
$$\sqrt{50}\sqrt{6}=5\sqrt{2}\cdot \sqrt{6}=5\sqrt{12}=10\sqrt{3},$$
so
$$\cos\theta=\frac{16}{10\sqrt{3}}=\frac{8\sqrt{3}}{15}.$$
Hence
$$\theta=\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right).$$